Tuesday, November 21, 2017

College Algebra, Chapter 4, 4.1, Section 4.1, Problem 10

A quadratic function f(x)=x2+8x.

a.) Find the quadratic function in standard form.


f(x)=x2+8xf(x)=(x2+8x+16)(1)(16)Complete the square: add (82)2=16 inside parentheses, subtract (1)(16) outsidef(x)=(x+4)216Factor and simplify


The standard form is f(x)=(x+4)216.

b.) Find its vertex and its x and y-intercepts.

Using the formula of standard form of a Quadratic function,

f(x)=a(xh)2+k

We know that the vertex is at (h,k), so the vertex of f(x)=(x+4)216 is at (4,16).


Solving for x-interceptsSolving for y-interceptWe set f(x)=0,thenWe set x=0, then0=(x+4)216Add 16y=(0+4)216Substitute x=016=(x+4)2Take the square rooty=1616Simplify±4=x+4Subtract 4y=0x=±44Simplifyx=0 and x=8


c.) Draw its graph.

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